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## SBN and IBN rings

Hello, I can not figure out why a ring that is not IBN (invariant basis number) must be SBN (single basis number). More precisely: Let $R$ be a ring (with unit, generally non-commutative) such that the free $R$-module $R^n$ is isomorphic to another free $R$-module $R^m$, where $n, m$ are different natural numbers. How does this imply that $R$ is isomorphic to $R^2$ as an $R$-module?

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 Are you sure that it is true? I am pretty sure that there are counterexamples. Also, this reminds me of mathoverflow.net/questions/10128/… – Martin Brandenburg Oct 8 2011 at 16:48 Oh, thank you, just now I have noticed your comment. I discussed it with someone else, but the idea for counterexample (see below) comes obviously from you. – Miroslav Korbelar Oct 8 2011 at 20:34

It seems that the implication does not hold. Thanks to Lukas Vokrinek for noticing this: According the example "Tom Leinster (mathoverflow.net/users/586), when is A isomorphic to A^3?" there is an abelian group $A$ such that $A$ is isomorphic to $A^3$ but not to $A^2$. Now, let $R=End(A)$. Then R is isomorphic to $R^3$ (as $R$-modules - that is easy), but not to $R^2$ (otherwise one could construct an isomorphism between $A$ and $A^2$ using the matrices with endomorphism entries, which would arise from the isomorphism $R$ and $R^2$).